Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

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You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 200000 + 10;
const int inf = 0x3f3f3f3f;
const long long mod = 7654321;
const double PI = 3.1415926;
const double eps = 1e-8;
int sy, a, b;
int n;
long long l[N], r[N];
int q;
long long sum[N];
int find2(double x) {
  int L = 1, R = n;
  int mid;
  int pos = 0;
  while (L <= R) {
    mid = (L + R) >> 1;
    if (l[mid] <= x) {
      L = mid + 1;
      pos = mid;
    } else
      R = mid - 1;
  }
  if (pos && r[pos] >= x) return pos;
  return 0;
}
int find(double x, int t) {
  int L = 1, R = n;
  int mid;
  int pos = 0;
  if (t == 0) {
    while (L <= R) {
      mid = (L + R) >> 1;
      if (l[mid] >= x) {
        R = mid - 1;
        pos = mid;
      } else
        L = mid + 1;
    }
  } else {
    while (L <= R) {
      mid = (L + R) >> 1;
      if (r[mid] <= x) {
        L = mid + 1;
        pos = mid;
      } else
        R = mid - 1;
    }
  }
  return pos;
}
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> sy >> a >> b;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> l[i] >> r[i];
  for (int i = 1; i <= n; i++) sum[i] = sum[i - 1] + r[i] - l[i];
  cin >> q;
  while (q--) {
    int x, y;
    cin >> x >> y;
    double ll2, rr;
    double s = 0;
    ll2 = (double)a + (double)(-sy) / ((-sy) + y) * (x - a);
    rr = (double)b + (double)(-sy) / ((-sy) + y) * (x - b);
    int pp1 = find2(ll2), pp2 = find2(rr);
    if (pp1 == pp2 && pp1) {
      double res = 0;
      res = (double)(rr - ll2) / y * ((-sy) + y);
      printf("%.9lf\n", res);
      continue;
    } else if (pp1 && pp2) {
      s += r[pp1] - ll2 + rr - l[pp2];
    } else if (pp1) {
      s += r[pp1] - ll2;
    } else if (pp2) {
      s += rr - l[pp2];
    }
    int p1, p2;
    p1 = find(ll2, 0), p2 = find(rr, 1);
    if (pp1) p1 = find(r[pp1], 0);
    if (pp2) p2 = find(l[pp2], 1);
    if (p1 && p2 && p1 <= p2) s += sum[p2] - (p1 ? sum[p1 - 1] : 0);
    double res = 0;
    res = (double)s / y * ((-sy) + y);
    printf("%.9lf\n", res);
  }
  return 0;
}